a m′

is therefore to

a m

as A C is to A B in [Fig. 79.]

Therefore P Q′ must be to P Q as A C is to A B.

But P Q equals P T ([Fig. 78.]); and P V is to V T (in [Fig. 78.]) as B E is to A E ([Fig. 79.]).

Hence we have only to substitute P V for E C, and V T for A E, in [Fig. 79.], and the resulting diagonal A C will be the required length of P Q′.

Fig. 80.

It will be seen that the construction given in the text ([Fig. 46.]) is the simplest means of obtaining this magnitude, for V D in [Fig. 46.] (or V M in [Fig. 15.]) = V T by construction in [Problem IV]. It should, however, be observed, that the distance P Q′ or P X, in [Fig. 46.], may be laid on the sight-line of the inclined plane itself, if the measuring-line be drawn parallel to that sight-line. And thus any form may be drawn on an inclined plane as conveniently as on a horizontal one, with the single exception of the radiation of the verticals, which have a vanishing-point, as shown in [Problem XX].